Problem 11 - Elementary Algebra

Question

Construct a mathematical model given the following. \(y\) varies directly as \(x,\) and \(y=30\) when \(x=6\).

Step-by-Step Solution

Answer

The mathematical model is \(y = 5x\).

Step 1: Understand Direct Variation

When we say that \(y\) varies directly as \(x\), it means there is a constant \(k\) such that \(y = kx\). This is called the direct variation formula.

Step 2: Set Up the Equation Using Given Values

We know \(y = 30\) when \(x = 6\). Plug these values into the direct variation equation: \(30 = k \times 6\).

Step 3: Solve for the Constant of Variation

To find the constant \(k\), solve the equation from Step 2: \(k = \frac{30}{6}\). Simplifying gives \(k = 5\).

Step 4: Write the Mathematical Model

Now that we know \(k = 5\), replace \(k\) in the direct variation formula to get the mathematical model: \(y = 5x\).

Key Concepts

Mathematical Model

A mathematical model is a representation of a real-world situation using mathematical concepts and language. In this exercise, we're creating a mathematical model to describe how two variables, \(y\) and \(x\), relate to each other.
The exercise tells us that \(y\) varies directly as \(x\). This suggests a straightforward relationship: as \(x\) changes, \(y\) changes at a constant rate.
To express this relationship mathematically, we use the direct variation formula \(y = kx\), where \(k\) is a constant.
This formula captures the essence of the problem, showing how one variable depends directly on another. It's a neat way to use mathematics to model situations where proportional changes occur between quantities.

Constant of Variation

The constant of variation, often denoted as \(k\), is a crucial part of direct variation problems. It represents the rate at which \(y\) changes with respect to \(x\).
In simpler words, \(k\) tells us how much \(y\) will increase when \(x\) increases by one unit.
In our example, we know that \(y = 30\) when \(x = 6\). By substituting these values into the direct variation equation \(y = kx\), we have \(30 = k \times 6\).
Solving for \(k\), we divide 30 by 6, and find that \(k = 5\).

• This means for every one unit increase in \(x\), \(y\) increases by 5 units due to the constant \(k=5\).

This constant defines our variation and is key to predicting \(y\) for any given \(x\).

Equation Solving

Equation solving is a fundamental mathematical skill. Here, it involves finding the unknown constant \(k\) in our direct variation model.
In practical terms, this means performing arithmetic operations to determine the value of \(k\) from known values of \(x\) and \(y\).
Start by setting up the equation with the given values: \(30 = k \times 6\). Next, reorganize the equation to solve for \(k\) by dividing both sides by 6. This gives \(k = \frac{30}{6}\), which simplifies to \(k = 5\).

• Solving such equations involves basic operations like multiplication, division, and simplification, which are standard methods in algebra.
• Getting comfortable with these operations empowers you to tackle more complex problems involving direct relationships.

Thus, the ability to solve equations is central to applying mathematical models successfully.